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I got asked the following question: Let $\{x_1,\dots ,x_r\}$ be a reduced system modulo $m$. For what intergers a and b the system $\{ax_1+b,\dots ,ax_r+b\}$ is also reduced modulo $m$?

Attempt:

well first its trivial that if $\{x_1,\dots ,x_r\}$ is reduced system modulo $m$ then if $\gcd(a,m)=1$ then $\{ax_1,\dots ,ax_r\}$ is also reduced system modulo $m$. Also I know that $x \to x+b$ preserves units mod $m$ iff $b$ is a multiple of rad$(m)$ as can be shown here. So the condition I found is $\gcd(a,m)=1$ and $b$ is a multiple of rad$(m)$.

I thought I was done but then someone suggested that if $m=6, a=2, b=3$, its clearly that $U_6=\{1,5\}$ (since $1$ and $5$ are the only numbers less than $6$ that are co-prime to $6$). so: $2\times1+3 = 5 \pmod 6, 5\times2+3 = 1 \pmod 6$ hence they remian a reduced system modulo m.

I am having a diffucult time understanding what other pattern is for $a$ and $b$. I thought after the exmaple I showed that we need $\gcd(a+b,m)=1$ but later I found a counter example.

Any help would be appreciated!

Eric Snyder
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Gardosh
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  • Edited for MathJax; you should take some time to look into how to work with it when asking (or answering) questions. The thing I'd suggest is to examine other low-count RRS (i.e., $\phi(m) < m/2$, as for $m=12,18$) and see what patterns follow. It's clear that $(a,m)=1$ should be sufficient for any prime modulus, but it's all those composite moduli that break that answer. – Eric Snyder Dec 09 '23 at 19:42
  • Also, it seems clear that by symmetry, if $(a,b)$ works, then $(-a,-b)$ works. For instance, for $m=6$ as above, $(4,3)$ works as well as $(2,3)$. Otherwise my paper scratchwork reveals nothing in the way of a pattern yet. :( – Eric Snyder Dec 09 '23 at 20:19

1 Answers1

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OK, from some pattern-searching, I have answers, but not proof. Below, $(a,b)$ are solutions as defined above.

First we note: For all moduli, $(a,b) = (x_i, 0)$ are valid solutions. In addition, for all prime moduli and all moduli $m = p_1p_2\cdots p_k$, these are the only valid solutions.

Now let's look case by case at the rest of the integers. We have three major cases and one minor case. In all cases, $j,k,\ell$ are natural numbers, and all $p$ are odd primes. Let $q = p_1^{j_1}p_2^{j_2} \cdots p_n^{j_n}$, and $r=p_1p_2 \cdots p_n$, that is, the product of the factors of $q$.

Case A: $m = q$. The additional solutions are $(a, b) = (x_i, kr)$, where $b<m$.

Case B: $m = 2q$. There are two sets of additional solutions. First, $(a, b) = (x_i, 2kr)$, where $b<m$. Second, $(a, b) = (2k, r + 2\ell r)$, for $b<m$ and $\gcd(a,m)=1$.

Case C: $m = 2^i q, i> 1$. Only the first set of additional solutions from Case B are valid, that is, $(a, b) = (x_i, 2kr)$, where $b<m$.

Case D: $m = p^j$. If there is only one prime factor in $q$, the same solution set as in case A exists: $(a, b) = (x_i, kp)$, where $b<m$. Importantly, this also holds for $m = 2^i$.

Eric Snyder
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